Question

$\int x \sin 3x \, dx$ is equal to _______.

• A. $\frac{1}{3}x + \frac{1}{9}\sin 3x + C$
• B. $\frac{1}{3}x \cos 3x + \frac{1}{9}\sin 3x + C$
• C. $-\frac{1}{3}x \cos 3x + \frac{1}{9}\sin 3x + C$
• D. $x \cos 3x - \sin 3x + C$

Answer 1

Answer: (C)

C

Answer 2

To evaluate the integral $\int x \sin 3x \, dx$, we use the method of integration by parts.

The formula for integration by parts is:

$$\int u \, dv = uv - \int v \, du$$

We need to choose $u$ and $dv$. According to the ILATE rule (Inverse, Logarithmic, Algebraic, Trigonometric, Exponential), we prioritize functions for $u$ in that order. Here, we have an algebraic function ($x$) and a trigonometric function ($\sin 3x$). 'Algebraic' comes before 'Trigonometric' in ILATE. So, we choose:

$u = x$ $dv = \sin 3x \, dx$

Now, we find $du$ by differentiating $u$, and $v$ by integrating $dv$:

1. Differentiate $u$:

$du = \frac{d}{dx}(x) \, dx = dx$

2. Integrate $dv$:

$v = \int \sin 3x \, dx$

To integrate $\sin 3x$, we can use a substitution (let $t = 3x$, then $dt = 3 \, dx \implies dx = \frac{1}{3} dt$).

$v = \int \sin t \left(\frac{1}{3}\right) dt = \frac{1}{3} \int \sin t \, dt = \frac{1}{3} (-\cos t) = -\frac{1}{3} \cos 3x$

Now, substitute $u$, $v$, and $du$ into the integration by parts formula:

$$\int x \sin 3x \, dx = (x)\left(-\frac{1}{3} \cos 3x\right) - \int \left(-\frac{1}{3} \cos 3x\right) \, dx$$ $$\int x \sin 3x \, dx = -\frac{1}{3} x \cos 3x + \frac{1}{3} \int \cos 3x \, dx$$

Next, we need to evaluate the remaining integral $\int \cos 3x \, dx$:

Similar to the previous integration, let $t = 3x$, then $dt = 3 \, dx \implies dx = \frac{1}{3} dt$.

$$\int \cos 3x \, dx = \int \cos t \left(\frac{1}{3}\right) dt = \frac{1}{3} \int \cos t \, dt = \frac{1}{3} \sin t = \frac{1}{3} \sin 3x$$

Substitute this back into our expression:

$$\int x \sin 3x \, dx = -\frac{1}{3} x \cos 3x + \frac{1}{3} \left(\frac{1}{3} \sin 3x\right) + C$$ $$\int x \sin 3x \, dx = -\frac{1}{3} x \cos 3x + \frac{1}{9} \sin 3x + C$$

Therefore, the final answer is $-\frac{1}{3} x \cos 3x + \frac{1}{9} \sin 3x + C$.